3 Templates for Binary Search

花花酱的lower_bound/upper_bound讲解

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**[disclaimer]以下内容整理自leetcode topics

Template I

int binarySearch(int[] nums, int target){
  if(nums == null || nums.length == 0)
    return -1;

  int left = 0, right = nums.length - 1;
  while(left <= right){
    // Prevent (left + right) overflow
    int mid = left + (right - left) / 2;
    if(nums[mid] == target){ return mid; }
    else if(nums[mid] < target) { left = mid + 1; }
    else { right = mid - 1; }
  }

  // End Condition: left > right
  return -1;
}

Template #1 is used to search for an element or condition which can be determined by accessing a single index in the array.

Key Attributes:

• Most basic and elementary form of Binary Search
• Search Condition can be determined without comparing to the element's neighbors (or use specific elements around it)
• No post-processing required because at each step, you are checking to see if the element has been found. If you reach the end, then you know the element is not found

Distinguishing Syntax:

• Initial Condition: left = 0, right = length-1
• Termination: left > right
• Searching Left: right = mid-1
• Searching Right: left = mid+1

Template II

int binarySearch(int[] nums, int target){
  if(nums == null || nums.length == 0)
    return -1;

  int left = 0, right = nums.length;
  while(left < right){
    // Prevent (left + right) overflow
    int mid = left + (right - left) / 2;
    if(nums[mid] == target){ return mid; }
    else if(nums[mid] < target) { left = mid + 1; }
    else { right = mid; }
  }

  // Post-processing:
  // End Condition: left == right
  if(left != nums.length && nums[left] == target) return left;
  return -1;
}

Template #2 is an advanced form of Binary Search. It is used to search for an element or condition which requires accessing the current index and its immediate right neighbor's index in the array.
**经验：l = mid + 1; r = mid;这个永远不会变，但是>=/<=/>/<这个要看情况。另外，r = length -1还是length也要看情况，这个的区别就是如果len是偶数,r = len + 1会取中间两个数右边那个。

Key Attributes:

• Search Condition needs to access element's immediate right neighbor
• Use element's right neighbor to determine if condition is met and decide whether to go left or right
• Gurantees Search Space is at least 2 in size at each step
• Post-processing required. Loop/Recursion ends when you have 1 element left. Need to assess if the remaining element meets the condition.

Distinguishing Syntax:

• Initial Condition: left = 0, right = length
• Termination: left == right
• Searching Left: right = mid
• Searching Right: left = mid+1

Template III

int binarySearch(int[] nums, int target) {
    if (nums == null || nums.length == 0)
        return -1;

    int left = 0, right = nums.length - 1;
    while (left + 1 < right){
        // Prevent (left + right) overflow
        int mid = left + (right - left) / 2;
        if (nums[mid] == target) {
            return mid;
        } else if (nums[mid] < target) {
            left = mid;
        } else {
            right = mid;
        }
    }

    // Post-processing:
    // End Condition: left + 1 == right
    if(nums[left] == target) return left;
    if(nums[right] == target) return right;
    return -1;
}

Template #3 is used to search for an element or condition which requires accessing the current index and its immediate left and right neighbor's index in the array.

Key Attributes:

• Search Condition needs to access element's immediate left and right neighbors
• Use element's neighbors to determine if condition is met and decide whether to go left or right
• Gurantees Search Space is at least 3 in size at each step
• Post-processing required. Loop/Recursion ends when you have 2 elements left. Need to assess if the remaining elements meet the condition.

Distinguishing Syntax:

• Initial Condition: left = 0, right = length-1
• Termination: left + 1 == right
• Searching Left: right = mid
• Searching Right: left = mid